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The matrix is $ M= \frac{d}{d\theta} e^{A+\theta B} \mid _{\theta = 0} $ where $A$ and $B$ are both $n\times n$ matrices. I was thinking solving it by introducing the equations: $\dot x = (A + \theta B)x,\ x(0) = I$ with solution $x = X(t,\theta) $, where $M = \frac{dX(1,0)}{d\theta}$, I was stuck then, thanks for any suggestions : )

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Welcome to Math.SE! Try the Taylor series for $\exp$: the term of 1st degree in $\theta$ should be exactly what you need. –  user53153 Jan 23 '13 at 1:37
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up vote 1 down vote accepted

To flesh out 5PMs comment: $$\exp(A + \theta B) = \sum_{k \geqslant 0} \frac{(A + \theta B)^k}{k!} = \sum_{k \geqslant 0} \frac{A^k + k A^{k-1}(\theta B) + \theta^2( \ldots )}{k!} $$Differentiating term by term, we get $$\sum_{k \geqslant 0} \frac{k A^{k-1} B + \theta( \ldots)}{k!}$$So when we evaluate at zero, we obtain: $$\sum_{k \geqslant 0} \frac{A^{k-1}}{(k-1)!} B = \exp(A)B$$

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Hmm, this works only when $A$ and $B$ commute. –  user1551 Jan 23 '13 at 13:24
    
hey man, yeah, I goofed that up. However, it nearly works, instead you get $\sum_{k \geqslant 0} \frac{1}{k!} (\sum_{j=0}^{k-1} A^j B A^{k-j})$ –  uncookedfalcon Jan 23 '13 at 20:38
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